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我是编程新手，我正在尝试用 Java 编写一个非常简单的 Rock、Paper、Scissors 游戏。它会编译并运行良好，但我想说“无效的移动。再试一次”。当用户（personPlay）没有输入正确的字符（r、p 或 s）时，或者类似的东西。最好的方法是什么？例如，如果你输入一个“q”，它应该打印“Invalid move”。非常感谢您！

// *************
// Rock.java 
// ************* 

import java.util.Scanner; 
import java.util.Random; 


public class Rock 
{ 
public static void main(String[] args) 
{ 
    String personPlay; //User's play -- "R", "P", or "S" 
    String computerPlay = ""; //Computer's play -- "R", "P", or "S" 
    int computerInt; //Randomly generated number used to determine 
                     //computer's play 
    String response; 


    Scanner scan = new Scanner(System.in); 
    Random generator = new Random(); 

    System.out.println("Hey, let's play Rock, Paper, Scissors!\n" + 
                       "Please enter a move.\n" + "Rock = R, Paper" + 
                       "= P, and Scissors = S.");

    System.out.println();

    //Generate computer's play (0,1,2) 
    computerInt = generator.nextInt(3)+1; 

    //Translate computer's randomly generated play to 
    //string using if //statements 

    if (computerInt == 1) 
       computerPlay = "R"; 
    else if (computerInt == 2) 
       computerPlay = "P"; 
    else if (computerInt == 3) 
       computerPlay = "S"; 


    //Get player's play from input-- note that this is 
    // stored as a string 
    System.out.println("Enter your play: "); 
    personPlay = scan.next();

    //Make player's play uppercase for ease of comparison 
    personPlay = personPlay.toUpperCase(); 

    //Print computer's play 
    System.out.println("Computer play is: " + computerPlay); 


    //See who won. Use nested ifs 

    if (personPlay.equals(computerPlay)) 
       System.out.println("It's a tie!"); 
    else if (personPlay.equals("R")) 
       if (computerPlay.equals("S")) 
          System.out.println("Rock crushes scissors. You win!!");
    else if (computerPlay.equals("P")) 
            System.out.println("Paper eats rock. You lose!!"); 
    else if (personPlay.equals("P")) 
       if (computerPlay.equals("S")) 
       System.out.println("Scissor cuts paper. You lose!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Paper eats rock. You win!!"); 
    else if (personPlay.equals("S")) 
         if (computerPlay.equals("P")) 
         System.out.println("Scissor cuts paper. You win!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Rock breaks scissors. You lose!!"); 
    else 
         System.out.println("Invalid user input."); 
}

}

4

5 回答 5

4

I would recommend making Rock, Paper and Scissors objects. The objects would have the logic of both translating to/from Strings and also "knowing" what beats what. The Java enum is perfect for this.

public enum Type{

  ROCK, PAPER, SCISSOR;

  public static Type parseType(String value){
     //if /else logic here to return either ROCK, PAPER or SCISSOR

     //if value is not either, you can return null
  }
}

The parseType method can return null if the String is not a valid type. And you code can check if the value is null and if so, print "invalid try again" and loop back to re-read the Scanner.

Type person=null;

 while(person==null){
      System.out.println("Enter your play: "); 
      person= Type.parseType(scan.next());
      if(person ==null){
         System.out.println("invalid try again");
      }
 }

Furthermore, your type enum can determine what beats what by having each Type object know:

public enum Type{

    //...

    //each type will implement this method differently
    public abstract boolean beats(Type other);


}

each type will implement this method differently to see what beats what:

ROCK{

   @Override
   public boolean beats(Type other){            
        return other ==  SCISSOR;

   }
}

 ...

Then in your code

 Type person, computer;
   if (person.equals(computer)) 
   System.out.println("It's a tie!");
  }else if(person.beats(computer)){
     System.out.println(person+ " beats " + computer + "You win!!"); 
  }else{
     System.out.println(computer + " beats " + person+ "You lose!!");
  }
于 2013-10-06T02:55:39.310 回答
0

Why not check for what the user entered and then ask the user to enter correct input again?

eg:

//Get player's play from input-- note that this is 
// stored as a string 
System.out.println("Enter your play: "); 
response = scan.next();
if(response=="R"||response=="P"||response=="S"){
  personPlay = response;
}else{
  System.out.println("Invaild Input")
}

for the other modifications, please check my total code at pastebin

于 2013-10-06T03:30:15.713 回答
0

if在我们尝试解决无效字符问题之前， and语句周围缺少花括号else if正在对您的程序逻辑造成严重破坏。将其更改为：

if (personPlay.equals(computerPlay)) {
   System.out.println("It's a tie!");
}
else if (personPlay.equals("R")) {
   if (computerPlay.equals("S")) 
      System.out.println("Rock crushes scissors. You win!!");
   else if (computerPlay.equals("P")) 
        System.out.println("Paper eats rock. You lose!!");
}
else if (personPlay.equals("P")) {
   if (computerPlay.equals("S")) 
       System.out.println("Scissor cuts paper. You lose!!"); 
   else if (computerPlay.equals("R")) 
        System.out.println("Paper eats rock. You win!!");
} 
else if (personPlay.equals("S")) {
     if (computerPlay.equals("P")) 
         System.out.println("Scissor cuts paper. You win!!"); 
     else if (computerPlay.equals("R")) 
        System.out.println("Rock breaks scissors. You lose!!");
}
else 
     System.out.println("Invalid user input.");

清晰多了！现在抓住坏角色实际上是小菜一碟。在尝试处理其他任何内容之前，您需要将else语句移动到可以捕获错误的位置。因此，将所有内容更改为：

if( /* insert your check for bad characters here */ ) { 
     System.out.println("Invalid user input.");
}
else if (personPlay.equals(computerPlay)) {
   System.out.println("It's a tie!");
}
else if (personPlay.equals("R")) {
   if (computerPlay.equals("S")) 
      System.out.println("Rock crushes scissors. You win!!");
   else if (computerPlay.equals("P")) 
        System.out.println("Paper eats rock. You lose!!");
}
else if (personPlay.equals("P")) {
   if (computerPlay.equals("S")) 
       System.out.println("Scissor cuts paper. You lose!!"); 
   else if (computerPlay.equals("R")) 
        System.out.println("Paper eats rock. You win!!");
} 
else if (personPlay.equals("S")) {
     if (computerPlay.equals("P")) 
         System.out.println("Scissor cuts paper. You win!!"); 
     else if (computerPlay.equals("R")) 
        System.out.println("Rock breaks scissors. You lose!!");
}
于 2013-10-06T03:15:33.340 回答
0 
int w =0 , l =0, d=0, i=0;
    Scanner sc = new Scanner(System.in);

// try tentimes
    while (i<10) {


        System.out.println("scissor(1) ,Rock(2),Paper(3) ");
        int n = sc.nextInt();
        int m =(int)(Math.random()*3+1);


        if(n==m){

            System.out.println("Com:"+m +"so>>> " + "draw");
            d++;


        }else if ((n-1)%3==(m%3)){
            w++;
            System.out.println("Com:"+m +"so>>> " +"win");
        }
        else if(n >=4 )
        {
            System.out.println("pleas enter correct number)");


    }
        else {
            System.out.println("Com:"+m +"so>>> " +"lose");
            l++;

        }
        i++;
于 2015-03-17T03:25:29.740 回答
0

你可以插入这样的东西：

personPlay = "B";

while (!personPlay.equals("R") && !personPlay.equals("P") && !personPlay.equals("S")) {

    //Get player's play from input-- note that this is 
    // stored as a string 
    System.out.println("Enter your play: "); 
    personPlay = scan.next();

    //Make player's play uppercase for ease of comparison 
    personPlay = personPlay.toUpperCase();

    if (!personPlay.equals("R") && !personPlay.equals("P") && !personPlay.equals("S"))
        System.out.println("Invalid move. Try again.");

}
于 2013-10-06T02:46:14.627 回答